Answer check: A ball on a spring launched into the air. find speed/height

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Homework Help Overview

The discussion revolves around a physics problem involving a ball launched by a spring, focusing on the conservation of mechanical energy to find the speed and height of the ball at various points in its trajectory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the conservation of energy equation, questioning the setup and assumptions regarding potential and kinetic energy at different points in the motion. There are discussions about the correct interpretation of the final position and the velocity at various points in the trajectory.

Discussion Status

Participants are actively comparing their calculations and results, with some expressing confusion over the correct application of energy conservation principles. There is a recognition of differing results and an ongoing exploration of the implications of their assumptions.

Contextual Notes

Some participants note the importance of identifying the correct point in the trajectory for calculating velocity, and there is mention of the spring's compression affecting the initial conditions of the problem.

joe426
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Homework Statement



aXvTc.png


Homework Equations



Since energy is conserved, MEinital = MEfinal

The Attempt at a Solution


1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2

After finding what equals 0, I am left with:
1/2kxi2 = 1/2mvf2 + mgyf
v=5.97m/s

I found yf by: mgyf = 1/2kxi2
yf = 3.63m


Just checking to see if these answers are correct. THank you for the help!
 
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joe426 said:
After finding what equals 0, I am left with:
1/2kxi2 = 1/2mvf2 + mgyf
v=5.97m/s
I get a different result. Please post more details of your calculation.
 
haruspex said:
I get a different result. Please post more details of your calculation.

First I found yf, which is the answer to part b.
Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with.

mgyf = 1/2kxi2
y= (1/2kxi2) / g
y=3.63m

Then,

1/2kxi2 = 1/2mvf2 + mgyf

sqrt((kxi2 - mgyf) / m) = v
v = 5.97m/s
 
Last edited:
joe426 said:
mgyf = 1/2kxi2
...
1/2kxi2 = 1/2mvf2 + mgyf
Compare those two equations. Shouldn't you get vf=0? ...
sqrt((kxi2 - mgyf) / m) = v
.. but you didn't because you dropped a factor of 2.
For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.
 
haruspex said:
Compare those two equations. Shouldn't you get vf=0? ...

.. but you didn't because you dropped a factor of 2.
For what point of the trajectory do you think you should be finding the velocity in part (a)? What is the value of y at that point.

oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air.

So its,
1/2mvi2 + 1/2kxi2 = mgyf
vi = sqrt( 2gyf - kxi2 )

But now I'm stuck not knowing what it's final position is.
 
Using yfinal = 3.63m. I got the velocity to be 7m/s
 
joe426 said:
Using yfinal = 3.63m. I got the velocity to be 7m/s

I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring

V= 8.3m/s for part a
Y= 3.6m for part b


Thanks for the help!
 

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